\(\int \frac {-15 x^4+e^x (-5+5 x)+(e^x (-1+x)-3 x^4) \log (\frac {e^x+80 x-x^4}{5 x})}{2 e^x x+160 x^2-2 x^5} \, dx\) [7726]

Optimal result

Integrand size = 69, antiderivative size = 27 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\frac {1}{4} \left (5+\log \left (16+\frac {e^x-x^4}{5 x}\right )\right )^2 \]

[Out]

1/2*(ln(16+1/5*(exp(x)-x^4)/x)+5)*(1/2*ln(16+1/5*(exp(x)-x^4)/x)+5/2)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6873, 12, 6818} \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\frac {1}{4} \left (\log \left (\frac {-x^4+80 x+e^x}{5 x}\right )+5\right )^2 \]

[In]

Int[(-15*x^4 + E^x*(-5 + 5*x) + (E^x*(-1 + x) - 3*x^4)*Log[(E^x + 80*x - x^4)/(5*x)])/(2*E^x*x + 160*x^2 - 2*x
^5),x]

[Out]

(5 + Log[(E^x + 80*x - x^4)/(5*x)])^2/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (e^x-e^x x+3 x^4\right ) \left (-5-\log \left (\frac {e^x+80 x-x^4}{5 x}\right )\right )}{2 x \left (e^x+80 x-x^4\right )} \, dx \\ & = \frac {1}{2} \int \frac {\left (e^x-e^x x+3 x^4\right ) \left (-5-\log \left (\frac {e^x+80 x-x^4}{5 x}\right )\right )}{x \left (e^x+80 x-x^4\right )} \, dx \\ & = \frac {1}{4} \left (5+\log \left (\frac {e^x+80 x-x^4}{5 x}\right )\right )^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\frac {1}{4} \left (5+\log \left (\frac {e^x+80 x-x^4}{5 x}\right )\right )^2 \]

[In]

Integrate[(-15*x^4 + E^x*(-5 + 5*x) + (E^x*(-1 + x) - 3*x^4)*Log[(E^x + 80*x - x^4)/(5*x)])/(2*E^x*x + 160*x^2
- 2*x^5),x]

[Out]

(5 + Log[(E^x + 80*x - x^4)/(5*x)])^2/4

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.83 (sec) , antiderivative size = 496, normalized size of antiderivative = 18.37

method	result	size
risch	\(\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{4}+\frac {i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}-x^{4}+80 x \right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2}}{4}-\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x}-x^{4}+80 x \right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{4}+\frac {i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}-x^{4}+80 x \right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{4}+\frac {i \ln \left (x \right ) \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{3}}{4}-\frac {i \ln \left (x \right ) \pi }{2}+\frac {\ln \left (5\right ) \ln \left (x \right )}{2}-\frac {5 \ln \left (x \right )}{2}+\frac {\ln \left (x \right )^{2}}{4}-\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x}-x^{4}+80 x \right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2}}{4}+\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{2}-\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{3}}{4}-\frac {i \pi \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2}}{2}+\frac {\ln \left (-{\mathrm e}^{x}+x^{4}-80 x \right )^{2}}{4}-\frac {i \ln \left (x \right ) \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{4}+\frac {i \ln \left (x \right ) \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{x}\right )}^{2}}{2}+\frac {5 \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{2}-\frac {\ln \left (x \right ) \ln \left (-{\mathrm e}^{x}+x^{4}-80 x \right )}{2}-\frac {\ln \left (5\right ) \ln \left ({\mathrm e}^{x}-x^{4}+80 x \right )}{2}\)	\(496\)

[In]

int((((-1+x)*exp(x)-3*x^4)*ln(1/5*(exp(x)-x^4+80*x)/x)+(5*x-5)*exp(x)-15*x^4)/(2*exp(x)*x-2*x^5+160*x^2),x,met
hod=_RETURNVERBOSE)

[Out]

1/4*I*Pi*ln(exp(x)-x^4+80*x)*csgn(I*(exp(x)-x^4+80*x)/x)^2*csgn(I/x)+1/4*I*ln(x)*Pi*csgn(I*(exp(x)-x^4+80*x))*
csgn(I*(exp(x)-x^4+80*x)/x)^2-1/4*I*Pi*ln(exp(x)-x^4+80*x)*csgn(I*(exp(x)-x^4+80*x))*csgn(I*(exp(x)-x^4+80*x)/
x)*csgn(I/x)+1/4*I*ln(x)*Pi*csgn(I*(exp(x)-x^4+80*x))*csgn(I*(exp(x)-x^4+80*x)/x)*csgn(I/x)+1/4*I*ln(x)*Pi*csg
n(I*(exp(x)-x^4+80*x)/x)^3-1/2*I*ln(x)*Pi+1/2*ln(5)*ln(x)-5/2*ln(x)+1/4*ln(x)^2-1/4*I*Pi*ln(exp(x)-x^4+80*x)*c
sgn(I*(exp(x)-x^4+80*x))*csgn(I*(exp(x)-x^4+80*x)/x)^2+1/2*I*Pi*ln(exp(x)-x^4+80*x)-1/4*I*Pi*ln(exp(x)-x^4+80*
x)*csgn(I*(exp(x)-x^4+80*x)/x)^3-1/2*I*Pi*ln(exp(x)-x^4+80*x)*csgn(I*(exp(x)-x^4+80*x)/x)^2+1/4*ln(-exp(x)+x^4
-80*x)^2-1/4*I*ln(x)*Pi*csgn(I*(exp(x)-x^4+80*x)/x)^2*csgn(I/x)+1/2*I*ln(x)*Pi*csgn(I*(exp(x)-x^4+80*x)/x)^2+5
/2*ln(exp(x)-x^4+80*x)-1/2*ln(x)*ln(-exp(x)+x^4-80*x)-1/2*ln(5)*ln(exp(x)-x^4+80*x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\frac {1}{4} \, \log \left (-\frac {x^{4} - 80 \, x - e^{x}}{5 \, x}\right )^{2} + \frac {5}{2} \, \log \left (-\frac {x^{4} - 80 \, x - e^{x}}{5 \, x}\right ) \]

[In]

integrate((((-1+x)*exp(x)-3*x^4)*log(1/5*(exp(x)-x^4+80*x)/x)+(5*x-5)*exp(x)-15*x^4)/(2*exp(x)*x-2*x^5+160*x^2
),x, algorithm="fricas")

[Out]

1/4*log(-1/5*(x^4 - 80*x - e^x)/x)^2 + 5/2*log(-1/5*(x^4 - 80*x - e^x)/x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).

Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=- \frac {5 \log {\left (x \right )}}{2} + \frac {\log {\left (\frac {- \frac {x^{4}}{5} + 16 x + \frac {e^{x}}{5}}{x} \right )}^{2}}{4} + \frac {5 \log {\left (- x^{4} + 80 x + e^{x} \right )}}{2} \]

[In]

integrate((((-1+x)*exp(x)-3*x**4)*ln(1/5*(exp(x)-x**4+80*x)/x)+(5*x-5)*exp(x)-15*x**4)/(2*exp(x)*x-2*x**5+160*
x**2),x)

[Out]

-5*log(x)/2 + log((-x**4/5 + 16*x + exp(x)/5)/x)**2/4 + 5*log(-x**4 + 80*x + exp(x))/2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).

Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=-\frac {1}{2} \, {\left (\log \left (5\right ) + \log \left (x\right ) - 5\right )} \log \left (-x^{4} + 80 \, x + e^{x}\right ) + \frac {1}{4} \, \log \left (-x^{4} + 80 \, x + e^{x}\right )^{2} + \frac {1}{2} \, {\left (\log \left (5\right ) - 5\right )} \log \left (x\right ) + \frac {1}{4} \, \log \left (x\right )^{2} \]

[In]

integrate((((-1+x)*exp(x)-3*x^4)*log(1/5*(exp(x)-x^4+80*x)/x)+(5*x-5)*exp(x)-15*x^4)/(2*exp(x)*x-2*x^5+160*x^2
),x, algorithm="maxima")

[Out]

-1/2*(log(5) + log(x) - 5)*log(-x^4 + 80*x + e^x) + 1/4*log(-x^4 + 80*x + e^x)^2 + 1/2*(log(5) - 5)*log(x) + 1
/4*log(x)^2

Giac [F]

\[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\int { \frac {15 \, x^{4} - 5 \, {\left (x - 1\right )} e^{x} + {\left (3 \, x^{4} - {\left (x - 1\right )} e^{x}\right )} \log \left (-\frac {x^{4} - 80 \, x - e^{x}}{5 \, x}\right )}{2 \, {\left (x^{5} - 80 \, x^{2} - x e^{x}\right )}} \,d x } \]

[In]

integrate((((-1+x)*exp(x)-3*x^4)*log(1/5*(exp(x)-x^4+80*x)/x)+(5*x-5)*exp(x)-15*x^4)/(2*exp(x)*x-2*x^5+160*x^2
),x, algorithm="giac")

[Out]

integrate(1/2*(15*x^4 - 5*(x - 1)*e^x + (3*x^4 - (x - 1)*e^x)*log(-1/5*(x^4 - 80*x - e^x)/x))/(x^5 - 80*x^2 -
x*e^x), x)

Mupad [B] (verification not implemented)

Time = 13.88 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {-15 x^4+e^x (-5+5 x)+\left (e^x (-1+x)-3 x^4\right ) \log \left (\frac {e^x+80 x-x^4}{5 x}\right )}{2 e^x x+160 x^2-2 x^5} \, dx=\frac {\ln \left (\frac {16\,x+\frac {{\mathrm {e}}^x}{5}-\frac {x^4}{5}}{x}\right )\,\left (\ln \left (\frac {16\,x+\frac {{\mathrm {e}}^x}{5}-\frac {x^4}{5}}{x}\right )+10\right )}{4} \]

[In]

int((log((16*x + exp(x)/5 - x^4/5)/x)*(exp(x)*(x - 1) - 3*x^4) + exp(x)*(5*x - 5) - 15*x^4)/(2*x*exp(x) + 160*
x^2 - 2*x^5),x)

[Out]

(log((16*x + exp(x)/5 - x^4/5)/x)*(log((16*x + exp(x)/5 - x^4/5)/x) + 10))/4